A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of $2\times10^5\ \text{m s}^{-1}$. When the electric field is switched off, the proton moves along a circular path of radius $2\ \text{cm}$. The magnitude of electric field is $x\times10^4\ \text{N/C}$. The value of $x$ is ______. Take the mass of the proton $= 1.6\times10^{-27}\ \text{kg}$.
Numerical value type. Enter your answer.
Answer: 2
With the magnetic field alone, $r = \dfrac{mv}{qB}$:
$$B = \frac{mv}{qr} = \frac{1.6\times10^{-27}\times2\times10^5}{1.6\times10^{-19}\times0.02} = 0.1\ \text{T}$$
Undeflected motion in crossed fields needs $qE = qvB$:
$$E = vB = 2\times10^5\times0.1 = 2\times10^4\ \text{N/C}$$
So $x = 2$.
Solution by Sreeraj P, M.Sc Physics